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Monica [59]
3 years ago
8

Give me lil reasoning so I know your not lying for points

Chemistry
1 answer:
ycow [4]3 years ago
4 0

Answer:

0.87 ATM

Explanation:

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Calculate the total quantity of heat required to convert 25.0 g of liquid ccl4(l from 35.0°c to gaseous ccl4 at 76.8°c (the norm
GREYUIT [131]

The solution would be like this for this specific problem:

<span>25.0 g CCl4 x (1 mole CCl4 / 153.8 g CCl4) = 0.163 moles CCl4 

 </span>(29.82 kJ / mole)(0.163 moles) = 4.86 kJ 

total heat = 1.11 kJ + 4.86 kJ = 5.97 kJ 

<span>I hope this helps and if you have any further questions, please don’t hesitate to ask again. </span>

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3 years ago
Which of the following statements accurately describes the structure of the atom?
Katarina [22]
The answer is b
Protons and neutrons are both in the nucleus
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5 0
3 years ago
50 points!! i need 3 answers!!!
tester [92]
Option E, Real gas particles have more complex interactions than ideal gas particles.

In ideal gases, there is absolutely no interaction between any atoms. At all. Atoms simply don't bump into each other in ideal gases.

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-T.B.
3 0
2 years ago
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Torrey's neighbor told her that a marble rolling down a hill increases in velocity as it rolls down, but does not increase in ki
ziro4ka [17]

Answer:

Torrey's neighbour is incorrect because increase in kinetic energy is proportional to velocity.  If the velocity increases so will the object's kinetic energy.  Because the mass is constant, if the velocity increases, so does the kinetic energy.

6 0
2 years ago
a compound has a molar mass of 129 g/mol if its empirical formula is C2H5N then what is the molecular formula
Elena-2011 [213]

Given :

A compound has a molar mass of 129 g/mol .

Empirical formula of compound is C₂H₅N .

To Find :

The molecular formula of the compound.

Solution :

Empirical mass of compound :

M_e = ( 2 \times 12 ) + ( 5 \times 1 ) + (  1  \times 14 )\\\\M_e = 43\ gram/mol

Now, n-factor is :

n = \dfrac{M}{M_e}\\\\n = \dfrac{129}{43}\\\\n = 3

Multiplying each atom in the formula by 3 , we get :

Molecular Formula, C₆H₁₅N₃

3 0
2 years ago
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